LET'S PUT IN WHAT WE KNOW!!!
Q
=
725 J
m
=
55.0 g
c
=
0.900 J/(°C⋅g)
Δ
T
=
final temperature - initial temperature
Δ
T
=
(
x
−
27.5
)
°C
We solve for
Δ
T
.
725 J
=
55.0 g
⋅
0.900 J/(°C⋅g)
(
x
−
27.5
)
°C
NOW IT'S JUST BASIC ALGREBRA
725
=
49.5
x
−
1361
2086
=
49.5
x
42.1
=
x
The final temperature is 42.1 °C.
Answer: water (H2O): Water is an excellent example of hydrogen bonding. ...
chloroform (CHCl3): Hydrogen bonding occurs between hydrogen of one molecule and carbon of another molecule.
ammonia (NH3): Hydrogen bonds form between hydrogen of one molecule and nitrogen of another.
Explanation:
Answer:
A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.
Explanation:
A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.
1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J
ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J
B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂
= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.
Answer:
botany because a maple tree is a plant
Explanation:
Answer:
E to F
Explanation:
We know that v = dx/dt
and a = d²x/dt²
Decreasing velocity means deceleration or negative acceleration,
therefore d²x/dt²<0
i.e the portion where the graph is having its curvature downwards like an inverted u, from graph E to F portion looks like that.
Therefore in the portion E to F velocity decreases.
